73 g of HCl and 8 g of O2 are passed over hot CuCl2 catalyst (Deacon process, 4HCl + O2 -> 2Cl2 + 2H2O, complete conversion of the limiting reagent). Taking H=1, O=16, Cl=35.5, the mass of Cl2 formed is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ph106

$73 g$ of $HCl$ and $8 g$ of $O X_{2}$ are passed over hot $CuCl X_{2}$ catalyst (Deacon process, $4 HCl + O X_{2} 2 Cl X_{2} + 2 H X_{2} O$ , complete conversion of the limiting reagent). Taking $H = 1$ , $O = 16$ , $Cl = 35.5$ , the mass of $Cl X_{2}$ formed is

Moles of

HCl = 73/36.5 = 2.0

and of

O X_{2} = 8/32 = 0.25

. Oxygen needs 4 mol HCl per mol, so

0.25

mol

O X_{2}

uses only 1.0 mol HCl, hence

O X_{2}

is limiting.

Cl X_{2}

formed

= 2 \times 0.25 = 0.50

mol, mass

= 0.50 \times 71 = 35.5 g